Answer:
longitudinal engineering strain = 624.16
true strain is 6.44
Explanation:
given data
diameter d1 = 0.5 mm
diameter d2 = 25 mm
to find out
longitudinal engineering and true strains
solution
we know both the volume is same
so
volume 1 = volume 2
A×L(1) = A×L(2)
( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)
( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)
0.1963 ×L(1) = 122.71 ×L(2)
L(1) / L(2) = 122.71 / 0.1963 = 625.16
and we know longitudinal engineering strain is
longitudinal engineering strain = L(1) / L(2) - 1
longitudinal engineering strain = 625.16 - 1
longitudinal engineering strain = 624.16
and
true strain is
true strain = ln ( L(1) / L(2))
true strain = ln ( 625.16)
true strain is 6.44
Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
(1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
(2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:
Next, you replace for all variables:
hence, the current in the wire is 4.75A
Given that,
Radius of track, r = 50 m
time , t = 9 s
velocity, v = ?
Distance covered by car in one lap around a track is equal to the circumference of the track.
C = 2 π r = 2 * 3.14 * 50
C = 314.159 m
Distance covered by car, s = 314.159 m
Velocity = distance/ time
V = 314.159 / 9
V = 34.9 m/s
The average velocity of car is 34.9 m/s.
Answer:
order d> a = e> c> b = f
Explanation:
Pascal's law states that a change in pressure is transmitted by a liquid, all points are transmitted regardless of the form
P₁ = P₂
Using the definition of pressure
F₁ / A₁ = F₂ / A₂
F₂ = A₂ /A₁ F₁
Now we can examine the results
a) F1 = 4.0 N A1 = 0.9 m2 A2 = 1.8 m2
F₂ = 1.8 / 0.9 4
F₂a = 8 N
b) F1 = 2.0 N A1 = 0.9 m2 A2 = 0.45 m2
F₂b = 0.45 / 0.9 2
F₂b = 1 N
c) F1 2.0 N A1 = 1.8 m2 A2 = 3.6 m2
F₂c = 3.6 / 1.8 2
F₂c = 4 N
d) F1 = 4.0N A1 = 0.45 m2 A2 = 1.8 m2
F₂d = 1.8 / 0.45 4.0
F₂d = 16 m2
e) F1 = 4.0 N A1 = 0.45 m2 A2 = 0.9 m2
F₂e = 0.9 / 0.45 4
F₂e = 8 N
f) F1 = 2.0N A1 = 1.8 m2 A2 = 0.9 m2
F₂f = 0.9 / 1.8 2.0
F₂f = 1 N
Let's classify the structure from highest to lowest
F₂d> F₂a = F₂e> F₂c> F₂b = F₂f
I mean the combinations are
d> a = e> c> b = f
<u>Answer:</u>
Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
Speed of truck = 25 m/s north = 25 j m/s
Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s
= (1.43 i + 1.00 j) m/s
Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j
Magnitude of velocity = 26.04 m/s
Angle from positive horizontal axis = 86.85⁰
So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.
